HUD-1686（kmp算法）

Oulipo

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18075    Accepted Submission(s): 7124

Problem Description

The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:

Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…

Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.

So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.

Input

The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:

One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.

 

Output

For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.

Sample Input

3

BAPC

BAPC

AZA

AZAZAZA

VERDI

AVERDXIVYERDIAN

 

Sample Output

1

3

0

刚开始直接就用了kmp，试了十几回，总是超时。以为要优化，后来发现是数组没开够。qwq

#include<stdio.h>
#include<string.h>
const int maxn=10006;
int nex[maxn];
char str[maxn*100],ptr[maxn];

void get_next(int n)
{
    int k=-1;
    nex[0]=-1;
    for(int i=1;i<n;i++){
        while(k>-1&&ptr[k+1]!=ptr[i])
            k=nex[k];
        if(ptr[k+1]==ptr[i])
            k++;
        nex[i]=k;
    }
}

int kmp(int n,int m)
{
    get_next(m);
    int ans=0,k=-1,i=0;
    for(;i<n;i++){
        while(k>-1&&str[i]!=ptr[k+1])
            k=nex[k];
        if(ptr[k+1]==str[i]){
            k++;
            if(k+1==m)
                ans++;
        }
    }
    return ans;
}

int main()
{
    int N;
    scanf("%d",&N);
    while(N--){
        getchar();
        scanf("%s%s",ptr,str);
        int n,m;
        n=strlen(str);
        m=strlen(ptr);
        printf("%d\n",kmp(n,m));
    }
    
    return 0;
}

有两种kmp。

#include<stdio.h>
#include<string.h>
const int maxn=10006;
int nex[maxn];
char str[maxn*100],ptr[maxn];

void get_next(int n)
{
    int k=-1;
    nex[0]=-1;
    for(int i=0;i<n;){
        if(k==-1||ptr[i]==ptr[k]){
            k++,i++;
            nex[i]=k;
        }
        else{
            k=nex[k];
        }
    }
}

int kmp(int n,int m)
{
    get_next(m);
    int ans=0,k=0,i=0;
    while(i<n){
        if(k==-1||ptr[k]==str[i]){
            ++k,++i;
            if(k==m){
                ++ans;
            }
        }
        else
            k=nex[k];
    }
    return ans;
}

int main()
{
    int N;
    scanf("%d",&N);
    while(N--){
        getchar();
        scanf("%s%s",ptr,str);
        int n,m;
        n=strlen(str);
        m=strlen(ptr);
        printf("%d\n",kmp(n,m));
    }
    
    return 0;
}